Calculate emf and internal resistance with two circuits with the same cell

[koalafacey]

Homework Statement

A cell is connected to a 10.0 Ohm resistor. The pd across the 10.0 Ohm resistor is 10.43 V.
A 4.00 Ohm resistor is connected in parallel with the 10.0 Ohm resistor. The pd across the 10.0 Ohm resistor falls to 7.87 V.
Calculate the emf and the internal resistance of the cell.

Homework Equations

E = V + Ir
V = IR

The Attempt at a Solution

I tried to work out the total resistance in circuit 2. I got it as 2.86 Ohms. I then tried to work out the current in this circuit and got it as 2.75A. Using the information of volts, current and resistance I used Ohm's Law to try and work out the value of the internal resistor. I got it as 2.85 Ohms. However, when i used this in circuit 1 and tried to work out Emf, the two circuits did not match. Help?

 

[cupid.callin]

a better way will be to use Kirchhoff's laws

you can find the current in 2 circuits using ohm's law and the potential drop across 2 resistor configuration.
then apply Kirchhoff loop law law and insert value of i in 2 circuits ... you'll get 2 eqn in 2 variables ... which you can solve

 

[koalafacey]

Thank you for the reply.
I'm doing AS physics and I haven't heard of Kirchhoff's law before.
Which equation would I use?

 

[cupid.callin]

Kirchhoff laws says that net potential drop/gain in any closed circuit is 0

for this ... give any loop a sense of rotation (clockwise or anticlockwise) and take potential drop as negative and potential gain as positive or vice verse

then list all potential drop and potential gain and equate them to 0

check this:
http://www.globalshiksha.com/content/examples-of-kirchhoffs-question
http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

 

[rwisz]

Based on the information provided, we can use the equations E = V + Ir and V = IR to calculate the emf and internal resistance of the cell. In circuit 1, the pd (voltage) across the 10.0 Ohm resistor is 10.43 V, and the resistance is 10.0 Ohms. Using V = IR, we can calculate the current to be 1.043 A. Plugging this into the equation E = V + Ir, we get E = 10.43 V + (1.043 A)(10.0 Ohms) = 20.43 V. Therefore, the emf of the cell is 20.43 V.

In circuit 2, the total resistance is 2.86 Ohms (calculated by adding the parallel resistances 10.0 Ohms and 4.00 Ohms). We already know the pd across the 10.0 Ohm resistor is 7.87 V. Using V = IR, we can calculate the current to be 0.787 A. Plugging this into the equation E = V + Ir, we get E = 7.87 V + (0.787 A)(2.86 Ohms) = 9.09 V. Therefore, the emf of the cell in circuit 2 is 9.09 V.

To calculate the internal resistance of the cell, we can use the equation E = V + Ir again, this time using the values we just calculated for emf and current in each circuit. In circuit 1, we get 20.43 V = 10.43 V + (1.043 A)(R), where R is the internal resistance. Solving for R, we get R = 10.0 Ohms. Similarly, in circuit 2 we get 9.09 V = 7.87 V + (0.787 A)(R), and solving for R gives us R = 2.86 Ohms. Therefore, the internal resistance of the cell is 2.86 Ohms.

It is important to note that the internal resistance of a cell is not a fixed value and can vary depending on factors such as age and temperature. Therefore, the values calculated in this problem may not be exact, but they provide a general understanding of how to calculate emf and internal resistance using the given information.